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3 years ago

GUYS please

Mathematics

1 answer:

Advocard [28]3 years ago

8 0

Answer:

2.64

Step-by-step explanation:

0.9(x + 1.4) - 2.3 + 0.1x = 1.6

0.9x + 1.26 -2.3 + 0.1x - 1.6 = 0

(0.9x + 0.1x) + (1.26 - 2.3 - 1.6) = 0

x - 2.64 = 0

x = 2.64

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30 POINS Find the measures of the numbered angles in the parallelogram. Please help

Whitepunk [10]

Answer:

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6 0

3 years ago

Hello there!

Usimov [2.4K]

firstly chain rule is dy/dx = dy/du * du/dx

a) y = 5u² + u - 1 hence dy/du = 10u +1

u = 3x +1 hence du/dx = 3

so dy/dx = 3(10u +1) {sub in u and expand}

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8 0

3 years ago

In a certain school district, it was observed that 32% of the students in the element schools were classified as only children (

kipiarov [429]

Answer:

We conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.

Step-by-step explanation:

We are given that in a certain school district, it was observed that 32% of the students in the element schools were classified as only children (no siblings).

However, in the special program for talented and gifted children, 135 out of 347 students are only children.

Let p = population proportion of only children in the special program.

So, Null Hypothesis, $H_0$ : p = 32% {means that the proportion of only children in the special program is equal to the proportion for the school district}

Alternate Hypothesis, $H_A$ : p $\neq$ 32% {means that the proportion of only children in the special program is significantly different from the proportion for the school district}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of only children in the special program = $\frac{135}{347}$ = 0.39

n = sample of students = 347

So, test statistics = $\frac{0.39-0.32}{\sqrt{\frac{0.39(1-0.39)}{347} } }$

= 2.673

The value of z test statistics is 2.673.

Now, at 0.02 significance level the z table gives critical value of -2.3263 and 2.3263 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.

4 0

3 years ago

Which is the graph of an odd monomial function?

Minchanka [31]

Odd function must pass through origin so first two options are not correct.

Odd function must be rotational symmetry around origin. So last option is not correct.

So answer is C.

6 0

4 years ago

The mean room temperature from Monday to Saturday is 25oC and the mean room temperature of Saturday and Sunday is 28oC. if the m

Thepotemich [5.8K]

Answer:

We can conclude that the room temperature on Saturday was 24°C

Step-by-step explanation:

For a set of N values:

{x₁, x₂, ..., xₙ}

The mean of the set is calculated as:

$M = \frac{x_1 + x_2 + ... + x_n}{N}$

In this case, our set is the temperature of 7 days (so we have 7 elements)

{T₁, T₂, T₃, T₄, T₅, T₆, T₇}

Such that:

T₆ = temperature on Saturday

T₇ = temperature on Sunday.

We know that:

"The mean room temperature from Monday to Saturday is 25°C"

Then:

$\frac{T_1 + T_2 + T_3 + T_4 + T_5 + T_6}{6} = 25 C$

"the mean room temperature of Saturday and Sunday is 28°C"

$\frac{T_6 + T_7}{2} = 28C$

"The mean room temperature from Monday to Sunday is 26°C"

$\frac{T_1 + T_2 + T_3 + T_4 + T_5 + T_6 + T_7}{7} = 26 C$

So we have 3 equations.

Let's rewrite:

T₁ + T₂ + T₃ + T₄ + T₅ = A

Then we can rewrite our equations as:

$\frac{A+ T_6}{6} = 25 C$

$\frac{T_6 + T_7}{2} = 28C$

$\frac{A + T_6 + T_7}{7} = 26 C$

Removing the "Celcius" and multiplying in the 3 equations by the denominator on both sides, we get:

A + T₆ = 6*25

T₆ + T₇ = 2*28

A + T₆ + T₇ = 7*26

We now need to solve that system for T₆

The first step is to isolate one of the variables in one of the equations, (because we want to solve this for T₆ , let's not isolate that one). Let's isolate A in the first one:

A = 6*25 - T₆

A = 150 - T₆

Now we can replace this on the other two equations:

T₆ + T₇ = 2*28

(150 - T₆ ) + T₆ + T₇ = 7*26

Now, let's isolate T₇ in the top equation to get:

T₇ = 2*28 - T₆

T₇ = 56 - T₆

Now we can replace this in the last equation to get:

(150 - T₆ ) + T₆ + (56 - T₆) = 7*26

Now, we can solve this for T₆

150 - T₆ + T₆ + 56 - T₆ = 182

-T₆ = 182 - 150 - 56

-T₆ = -24

T₆ = 24

We can conclude that the room temperature on Saturday was 24°C

7 0

3 years ago